Re: [R] Elegant way to subtract matrix from array
A-kronecker(rep(1,10),S)
kronecker(m1,m2) creates a "tiled" matrix
each element of m1 in replaced by m2 multiplied with the element of m1
m1 = (1 2)
(3 4)
m2 = (11 12)
(13 14)
kronecker(m1,m2) therefore is
1 * (11 12)2 * (11 12)
(13 14)(13 14)
3 * (11 12)3 * (11 12)
(13 14)(13 14)
11 13 22 26
12 14 24 28
33 39 44 52
36 42 48 56
If m1 has 1 everywhere, the tiles are all identical.
On 7/27/2011 11:31 AM, steven mosher wrote:
> Cool,
>
> I looked at sweep but didnt consider it as I thought it was restricted to
> certain functions.
> So thanks for that solution.
>
> yes the data is very large and the future work will increase 10 fold,
>
> as for the matrix one I'm not too keen on replicating the smaller matrix,
> I've had one guy using the package who has hit the memory limits.. I have
> one more thing to try
>
> Thanks!
>
> Steve
>
> On Wed, Jul 27, 2011 at 1:42 AM, Gavin Simpson wrote:
>
>> On Wed, 2011-07-27 at 01:06 -0700, steven mosher wrote:
>>> there are really two related problems here
>>>
>>> I have a 2D matrix
>>>
>>>
>>> A <- matrix(1:100,nrow=20,ncol =5)
>>>
>>>
>>> S <- matrix(1:10,nrow=2,ncol =5)
>>>
>>>
>>> #I want to subtract S from A. so that S would be subtracted from the
>>> first 2 rows of
>>>
>>> #A, then the next two rows and so on.
>>
>> For this one, I have used the following trick to replication a matrix
>>
>>do.call(rbind, rep(list(mat), N)
>>
>> where we convert the matrix, `mat`, to a list and repeat that list `N`
>> times, and arrange for the resulting list to be rbind-ed. For your
>> example matrices, the following does what you want:
>>
>>A - do.call(rbind, rep(list(S), nrow(A)/nrow(S)))
>>
>> Whether this is useful will depend on the dimension of A and S - from
>> your posts on R-Bloggers, I can well imagine you are dealing with large
>> matrices.
>>
>>> #I have a the same problem with a 3D array
>>>
>>> # where I want to subtract Q for every layer (1-10) in Z
>>>
>>> # I thought I solved this one with array(mapply("-",Z,Q),dim=dim(Z))
>>>
>>> # but got the wrong answers
>>>
>>>
>>> Z <- array(1:100,dim=c(2,5,10))
>>>
>>> Q <- matrix(1:10,nrow=2,ncol =5)
>>
>> For this one, consider the often overlooked function `sweep()`:
>>
>>sweep(Z, c(1,2), Q, "-")
>>
>> does what you wanted. c(1,2) is the `MARGIN` argument over the
>> dimensions that Q will be swept from.
>>
>> HTH
>>
>> G
>>
>> --
>> %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%
>> Dr. Gavin Simpson [t] +44 (0)20 7679 0522
>> ECRC, UCL Geography, [f] +44 (0)20 7679 0565
>> Pearson Building, [e] gavin.simpsonATNOSPAMucl.ac.uk
>> Gower Street, London [w] http://www.ucl.ac.uk/~ucfagls/
>> UK. WC1E 6BT. [w] http://www.freshwaters.org.uk
>> %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%
>>
>>
>
> [[alternative HTML version deleted]]
>
> __
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>
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Re: [R] Elegant way to subtract matrix from array
Thanks Gabor!
On Wed, Jul 27, 2011 at 3:08 AM, Gabor Grothendieck wrote:
> On Wed, Jul 27, 2011 at 4:06 AM, steven mosher
> wrote:
> > there are really two related problems here
> >
> > I have a 2D matrix
> >
> >
> > A <- matrix(1:100,nrow=20,ncol =5)
> >
> >
> > S <- matrix(1:10,nrow=2,ncol =5)
> >
> >
> > #I want to subtract S from A. so that S would be subtracted from the
> > first 2 rows of
> >
> > #A, then the next two rows and so on.
> >
> >
> > #I have a the same problem with a 3D array
> >
> > # where I want to subtract Q for every layer (1-10) in Z
> >
> > # I thought I solved this one with array(mapply("-",Z,Q),dim=dim(Z))
> >
> > # but got the wrong answers
> >
> >
> > Z <- array(1:100,dim=c(2,5,10))
> >
> > Q <- matrix(1:10,nrow=2,ncol =5)
> >
>
> For the first one:
>
> matrix(c(t(A)) - c(t(S)), nrow(A), byrow = TRUE)
>
> or this version which may seem a bit more complex but has the
> advantage that it shows the general form of which both your questions
> are special cases:
>
>ix <- 2:1
>aperm(array(c(aperm(A, ix)) - c(t(S)), dim(A)[ix]), ix)
>
> Now, as mentioned, the answer to second question is the same except for ix:
>
>ix <- c(2, 1, 3)
>aperm(array(c(aperm(Z, ix)) - c(t(Q)), dim(Z)[ix]), ix)
>
>
> --
> Statistics & Software Consulting
> GKX Group, GKX Associates Inc.
> tel: 1-877-GKX-GROUP
> email: ggrothendieck at gmail.com
>
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Re: [R] Elegant way to subtract matrix from array
On Wed, Jul 27, 2011 at 4:06 AM, steven mosher wrote:
> there are really two related problems here
>
> I have a 2D matrix
>
>
> A <- matrix(1:100,nrow=20,ncol =5)
>
>
> S <- matrix(1:10,nrow=2,ncol =5)
>
>
> #I want to subtract S from A. so that S would be subtracted from the
> first 2 rows of
>
> #A, then the next two rows and so on.
>
>
> #I have a the same problem with a 3D array
>
> # where I want to subtract Q for every layer (1-10) in Z
>
> # I thought I solved this one with array(mapply("-",Z,Q),dim=dim(Z))
>
> # but got the wrong answers
>
>
> Z <- array(1:100,dim=c(2,5,10))
>
> Q <- matrix(1:10,nrow=2,ncol =5)
>
For the first one:
matrix(c(t(A)) - c(t(S)), nrow(A), byrow = TRUE)
or this version which may seem a bit more complex but has the
advantage that it shows the general form of which both your questions
are special cases:
ix <- 2:1
aperm(array(c(aperm(A, ix)) - c(t(S)), dim(A)[ix]), ix)
Now, as mentioned, the answer to second question is the same except for ix:
ix <- c(2, 1, 3)
aperm(array(c(aperm(Z, ix)) - c(t(Q)), dim(Z)[ix]), ix)
--
Statistics & Software Consulting
GKX Group, GKX Associates Inc.
tel: 1-877-GKX-GROUP
email: ggrothendieck at gmail.com
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Elegant way to subtract matrix from array
Cool,
I looked at sweep but didnt consider it as I thought it was restricted to
certain functions.
So thanks for that solution.
yes the data is very large and the future work will increase 10 fold,
as for the matrix one I'm not too keen on replicating the smaller matrix,
I've had one guy using the package who has hit the memory limits.. I have
one more thing to try
Thanks!
Steve
On Wed, Jul 27, 2011 at 1:42 AM, Gavin Simpson wrote:
> On Wed, 2011-07-27 at 01:06 -0700, steven mosher wrote:
> > there are really two related problems here
> >
> > I have a 2D matrix
> >
> >
> > A <- matrix(1:100,nrow=20,ncol =5)
> >
> >
> > S <- matrix(1:10,nrow=2,ncol =5)
> >
> >
> > #I want to subtract S from A. so that S would be subtracted from the
> > first 2 rows of
> >
> > #A, then the next two rows and so on.
>
> For this one, I have used the following trick to replication a matrix
>
>do.call(rbind, rep(list(mat), N)
>
> where we convert the matrix, `mat`, to a list and repeat that list `N`
> times, and arrange for the resulting list to be rbind-ed. For your
> example matrices, the following does what you want:
>
>A - do.call(rbind, rep(list(S), nrow(A)/nrow(S)))
>
> Whether this is useful will depend on the dimension of A and S - from
> your posts on R-Bloggers, I can well imagine you are dealing with large
> matrices.
>
> > #I have a the same problem with a 3D array
> >
> > # where I want to subtract Q for every layer (1-10) in Z
> >
> > # I thought I solved this one with array(mapply("-",Z,Q),dim=dim(Z))
> >
> > # but got the wrong answers
> >
> >
> > Z <- array(1:100,dim=c(2,5,10))
> >
> > Q <- matrix(1:10,nrow=2,ncol =5)
>
> For this one, consider the often overlooked function `sweep()`:
>
>sweep(Z, c(1,2), Q, "-")
>
> does what you wanted. c(1,2) is the `MARGIN` argument over the
> dimensions that Q will be swept from.
>
> HTH
>
> G
>
> --
> %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%
> Dr. Gavin Simpson [t] +44 (0)20 7679 0522
> ECRC, UCL Geography, [f] +44 (0)20 7679 0565
> Pearson Building, [e] gavin.simpsonATNOSPAMucl.ac.uk
> Gower Street, London [w] http://www.ucl.ac.uk/~ucfagls/
> UK. WC1E 6BT. [w] http://www.freshwaters.org.uk
> %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%
>
>
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R-help@r-project.org mailing list
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Re: [R] Elegant way to subtract matrix from array
On Wed, 2011-07-27 at 01:06 -0700, steven mosher wrote:
> there are really two related problems here
>
> I have a 2D matrix
>
>
> A <- matrix(1:100,nrow=20,ncol =5)
>
>
> S <- matrix(1:10,nrow=2,ncol =5)
>
>
> #I want to subtract S from A. so that S would be subtracted from the
> first 2 rows of
>
> #A, then the next two rows and so on.
For this one, I have used the following trick to replication a matrix
do.call(rbind, rep(list(mat), N)
where we convert the matrix, `mat`, to a list and repeat that list `N`
times, and arrange for the resulting list to be rbind-ed. For your
example matrices, the following does what you want:
A - do.call(rbind, rep(list(S), nrow(A)/nrow(S)))
Whether this is useful will depend on the dimension of A and S - from
your posts on R-Bloggers, I can well imagine you are dealing with large
matrices.
> #I have a the same problem with a 3D array
>
> # where I want to subtract Q for every layer (1-10) in Z
>
> # I thought I solved this one with array(mapply("-",Z,Q),dim=dim(Z))
>
> # but got the wrong answers
>
>
> Z <- array(1:100,dim=c(2,5,10))
>
> Q <- matrix(1:10,nrow=2,ncol =5)
For this one, consider the often overlooked function `sweep()`:
sweep(Z, c(1,2), Q, "-")
does what you wanted. c(1,2) is the `MARGIN` argument over the
dimensions that Q will be swept from.
HTH
G
--
%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%
Dr. Gavin Simpson [t] +44 (0)20 7679 0522
ECRC, UCL Geography, [f] +44 (0)20 7679 0565
Pearson Building, [e] gavin.simpsonATNOSPAMucl.ac.uk
Gower Street, London [w] http://www.ucl.ac.uk/~ucfagls/
UK. WC1E 6BT. [w] http://www.freshwaters.org.uk
%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%
__
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Elegant way to subtract matrix from array
there are really two related problems here
I have a 2D matrix
A <- matrix(1:100,nrow=20,ncol =5)
S <- matrix(1:10,nrow=2,ncol =5)
#I want to subtract S from A. so that S would be subtracted from the
first 2 rows of
#A, then the next two rows and so on.
#I have a the same problem with a 3D array
# where I want to subtract Q for every layer (1-10) in Z
# I thought I solved this one with array(mapply("-",Z,Q),dim=dim(Z))
# but got the wrong answers
Z <- array(1:100,dim=c(2,5,10))
Q <- matrix(1:10,nrow=2,ncol =5)
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
